Ampere's Law for Cylindrical Conductor

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amwil
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Homework Statement:
A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 6.85 cm and inner radius Rb = 3.75 cm . (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.85 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 4.74 cm from the axis of the conducting tube?
What is the expression for the current Iencl enclosed in the path of integration in terms of the current I , the outer radius Ra , the inner radius Rb , and the distance from the axis r where Ra>r>Rb ?
Express your answer in terms of I , Ra , Rb , and r .
Relevant Equations:
I = JA
I know that Ienl for the inner cylinder is just I and the current density for the outer tube is J1= -I/(pi(Ra^2-Rb^2). I assume that the current through the enclosed portion of the conducting tube (I1) is equal to J1(A1) where A1 is the area of the enclosed portion of the conducting tube. I found A1=pi(Ra^2-r^2) then multiplied it by J1 to get I1 = (-I(Ra^2-r^2))/(Ra^2-Rb^2). Then I added Iencl for the inner cylinder and I1 for the outer tube to get Iencl= I + (-I(Ra^2-r^2))/(Ra^2-Rb^2) but it said this was wrong. Can someone tell me what I'm doing wrong??
 

Answers and Replies

  • #2
TSny
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I know that Ienl for the inner cylinder is just I and the current density for the outer tube is J1= -I/(pi(Ra^2-Rb^2).
OK

I assume that the current through the enclosed portion of the conducting tube (I1) is equal to J1(A1) where A1 is the area of the enclosed portion of the conducting tube.
OK

I found A1=pi(Ra^2-r^2)
Are you sure this is the correct expression for the area of the enclosed portion of the tube?
 
  • #3
SammyS
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To be more explicit: Is ##R_a## the inner radius or is it the outer radius of the tube?
 

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